It is known that if a manifold is stably parallelizable, then it's Stiefel-Whitney classes must vanish. Is the converse true?

Note that we know that the converse cannot hold if stably parallelizable is replaced parallelizable; for example, spheres are stably parallelizable and so have vanishing Stiefel-Whitney classes but are not parallelizable. Also, what is known about this question if we replace Stiefel-Whitney classes with other characteristic classes?

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    $\begingroup$ [Sutherland, W. A. "Vanishing of Stiefel-Whitney classes." Proceedings of the American Mathematical Society 31.2 (1972): 637.] notes that Atiyah and Hirzebruch have shown that S-W classes vanish for all vector bundles on the 9th suspension of every finite CW-complex. $\endgroup$ – Mariano Suárez-Álvarez Dec 7 '14 at 9:21
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    $\begingroup$ @Mariano: but when is the 9th suspension of a finite CW complex both 1) a manifold and 2) not stably parallelizable? $\endgroup$ – Qiaochu Yuan Dec 7 '14 at 20:49
  • $\begingroup$ Related: math.stackexchange.com/questions/46297/… $\endgroup$ – Grigory M Dec 25 '14 at 15:42
  • $\begingroup$ An example of a non-stably trivial vector bundle with trivial Stiefel-Whitney and Pontryagin classes is given here. $\endgroup$ – Michael Albanese May 28 '17 at 14:13

The converse is not true. If a manifold is stably parallelizable, then its Pontryagin classes must also vanish, but the vanishing of the Stiefel-Whitney classes need not imply this.

For example, let $X$ be a closed simply connected smooth $4$-manifold. Since $X$ is simply connected, $w_1$ and $w_3$ automatically vanish. $w_4$ vanishes iff the Euler characteristic of $X$ is even iff $\dim H_2(X)$ is even. $w_2$ vanishes iff $X$ admits a spin structure iff the intersection form of $X$ is even. It follows that the Stiefel-Whitney classes of $X$ vanish iff its intersection form has even rank and even parity. But by the Hirzebruch signature theorem, the Pontryagin class $p_1 \in H^4(X)$ vanishes iff the signature $\sigma(X)$ of the intersection form vanishes.

Hence to give a counterexample it suffices to find a closed simply connected smooth $4$-manifold $X$ whose intersection form has even rank, even parity, and nonzero signature. For example, you can take $X$ to be a K3 surface, or more explicitly a hypersurface of degree $4$ in $\mathbb{CP}^3$ (see, for example, the calculations in this blog post), whose intersection form has rank $22$, even parity, and signature $-16$. By Rokhlin's theorem this is the smallest signature possible (in absolute value).

Edit: Incidentally, $4$ is the smallest possible dimension of a closed counterexample. If $X$ is a closed smooth $d$-manifold, $d \le 3$, then:

  • When $d = 1$, $X$ is a disjoint union of circles and hence is parallelizable.
  • When $d = 2$, if $w_1$ vanishes then $X$ is orientable, and this automatically implies that $w_2$ vanishes (e.g. because we know that the Euler characteristic is even in this case, but see this blog post for other arguments). But a closed orientable surface is stably parallelizable: in the usual embedding of such a surface into $\mathbb{R}^3$, the outward pointing normal trivializes the normal bundle.
  • When $d = 3$, if $w_1$ vanishes then $X$ is orientable, and again this automatically implies that $w_2$ vanishes, now by some Wu class computations. It follows that the classifying map of the tangent bundle $X \to BO(3)$ lifts to a map $X \to B \text{Spin}(3)$. But $B \text{Spin}(3)$ is $3$-connected, so any map to it from a $3$-manifold is nullhomotopic. Hence $X$ is not only stably parallelizable but parallelizable.
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    $\begingroup$ Thanks! What about when both the Stiefel-Whitney and the Pontryagin numbers vanish? Then the manifold must be null-cobordant but does this imply that the manifold is stably parallelizable? $\endgroup$ – user39598 Dec 7 '14 at 22:02
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    $\begingroup$ @user39598: you need to be more careful with your wording here. The condition that characteristic numbers vanish is weaker than the condition that characteristic classes vanish, although the latter implies the former. And you should distinguish between being null-cobordant and being oriented null-cobordant. Anyway, I believe the answer is no, but I don't have a counterexample. $\endgroup$ – Qiaochu Yuan Dec 7 '14 at 22:05
  • $\begingroup$ Here is a strategy for constructing a counterexample: suppose a smooth manifold $X$ admits a stable spin structure, or equivalently that the classifying map $X \to BO$ of its stable tangent bundle lifts to a map $X \to B \text{Spin}$. Now, $B \text{Spin}$ is $3$-connected with $\pi_4 \cong \mathbb{Z}$. There is a natural map $B \text{Spin} \to B^4 \mathbb{Z}$ inducing an isomorphism on $\pi_4$, and it gives a characteristic class of stable spin structures called the first fractional Pontryagin class $\frac{p_1}{2}$, so named because twice the class is the pullback of $p_1$. $\endgroup$ – Qiaochu Yuan Dec 7 '14 at 22:07
  • $\begingroup$ This characteristic class vanishes iff $X \to B \text{Spin}$ lifts to the next step in the Whitehead tower (ncatlab.org/nlab/show/Whitehead+tower) of $BO$, which is called $B \text{String}$ (see ncatlab.org/nlab/show/string+structure for details). In particular, if $H^4(X, \mathbb{Z})$ has $2$-torsion (so $\dim X \ge 5$) then it's possible that $p_1$ vanishes but that $\frac{p_1}{2}$ does not (and we can ignore the other Pontryagin classes if $\dim X \le 7$). But I can't complete the construction of a counterexample from here because I don't know how to compute $\frac{p_1}{2}$. $\endgroup$ – Qiaochu Yuan Dec 7 '14 at 22:08
  • $\begingroup$ Incidentally, this argument shows that if $X$ is a closed smooth $4$-manifold whose Stiefel-Whitney and Pontryagin classes vanish, then $X$ is stably parallelizable. But I don't know what happens when $\dim X \ge 5$. $\endgroup$ – Qiaochu Yuan Dec 7 '14 at 22:12

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